(4/3)x+3=23/3

Solution for (4/3)x+3=23/3 equation:

(4/3)x+3=23/3

We move all terms to the left:

(4/3)x+3-(23/3)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+4/3)x+3-(+23/3)=0

We multiply parentheses

4x^2+3-(+23/3)=0

We get rid of parentheses

4x^2+3-23/3=0

We multiply all the terms by the denominator


4x^2*3-23+3*3=0

We add all the numbers together, and all the variables

4x^2*3-14=0

Wy multiply elements

12x^2-14=0

a = 12; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·12·(-14)
Δ = 672
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{42}}{2*12}=\frac{0-4\sqrt{42}}{24}
=-\frac{4\sqrt{42}}{24}
=-\frac{\sqrt{42}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{42}}{2*12}=\frac{0+4\sqrt{42}}{24}
=\frac{4\sqrt{42}}{24}
=\frac{\sqrt{42}}{6}
$